Q7.3 - 6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1-20, determine the Laplace transform of the given function using Table 7.1 on page 356 and the properties of the transform given in Table 7.2. [Hint: In Problems 12-20, use an appropriate trigonometric identity.]

${\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\mathbf{2}\mathbf{t}\mathbf{+}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}{\mathbf{t}}^{\mathbf{2}}$

Step-by-Step Solution

Answer

The Laplace transform for the given equation is $\frac{2}{{(s+2)}^2+4}+\frac{2}{{(s-3)}^3}$.

Definition of Laplace transform

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transform. 
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by $\mathcal{L}\left\{f\left(t\right)\right\}$ or F(s).

Determine the Laplace transform for the given equation

Given that $e^{-2t}\sin 2t+e^{3t}{t}^2$.

Find the Laplace transform of the given function $e^{-2t}\sin 2t+e^{3t}{t}^2$ using the Laplace formula $\mathcal{L}\left\{a{f}_1+b{f}_2\right\}=a\mathcal{L}\left\{f_1\right\}+b\mathcal{L}\left\{f_2\right\}$, $\mathcal{L}\left\{e^{at}{t}^n\right\}=\frac{n!}{{(s-a)}^{n+1}}$ and $\mathcal{L}\left\{e^{at}sinbt\right\}=\frac{b}{{(s-a)}^2+b^2}$ as follows:

$\begin{array}{rcl}\mathcal{L}\left\{e^{-2t}\sin 2t+e^{3t}{t}^2\right\}& =& \mathcal{L}\left\{e^{-2t}\sin 2t\right\}+\mathcal{L}\left\{e^{3t}{t}^2\right\}\\ & =& \frac{2}{{(s-(-2\left)\right)}^2+2^2}+\frac{2!}{{(s-3)}^{2+1}}\\ & =& \frac{2}{{(s+2)}^2+4}+\frac{2}{{(s-3)}^3}\end{array}$

Therefore, the Laplace transform for the given equation is $\frac{2}{{(s+2)}^2+4}+\frac{2}{{(s-3)}^3}$.